Evaluate integral 1-cos^2 x/sin^2 x upper sum is pi and lower sum is 0?

Answer 1

is it 1-(cos^2x/sin^2x) or (1-cos^x)/sin^2x ?

Answer 2

If you mean (1 - cos^2(x)) / sin^2(x), and not 1 - (cos^2(x) / sin^2(x)), then since sin^2(x) + cos^2(x) = 1, this simplifies to sin^2(x) / sin^2(x) = 1. This gives you x when you integrate, so the answer is pi - 0 = pi.

Note that the original expression is undefined at 0 and pi. However, the function is still continuous up to there.

Answer 3

Great confusion , as can be seen from the answers , due to lack of brackets.
Will hazard a guess (and it can only be a guess) at:-
I = ∫ (1 - cos ² x) / sin ² x dx
I = ∫ sin ² x / sin ² x dx
I = ∫ 1 dx between limits of 0 and π
I = x
I = π
You might be good enough to let us know what the original question is meant to be.

Answer 4

∫1 - cos^2 x/sin^2 x dx
= ∫2 - csc^2 x dx
= 2x + cotx, x from 0 to pi =>undefined

∫(1 - cos^2 x)/sin^2 x dx
= ∫dx, x from 0 to pi
= pi

Answer 5

f(x) = ?(0 to pi, (a million + cos(x)) dx ) First, evaluate the necessary. The necessary of one million is x, and the necessary of cos(x) is sin(x). Defining F(x) to be the necessary, F(x) = x + sin(x), in an attempt to judge our convinced necessary: F(pi) - F(0) [ pi - sin(pi) ] - [ 0 - sin(0) ] [ pi - 0 ] - [ 0 - 0 ] pi - 0 pi

Answer 6

use this trig identity:

sin [x] ^2 + cos [x]^2 = 1

this implies 1 - cos [x]^2 = sin [x]^2

your new integral is

int (pi to 0) sin [x]^2/ sin [x]^2 dx

= int (pi to 0) 1 dx

= x

sub in your values

= Pi - 0
= Pi

Answer 7

I assume you mean 1 - (cos²x/sin²x), otherwise the answer is trivial.

∫[1 - (cos²x/sin²x)]dx = ∫[1 - cot²x]dx

= x + (cot x + x) = 2x + cot x

However, cot x is undefined at both x = 0 and x = π.

Answer 8

If the problem states:

Integral (1-cos^2x) / sin^2x dx

Integral sin^2x / sin^2x dx

Integral dx = x

Answer: pi

If the problem states:

Integral 1 - (cos^2x / sin^2x) dx

Integral 1 - tan^2 x dx

x - (Integral tan^2x dx)

(Integral tan^2x dx)

x - (tan x - x)

Answer: 2pi - tan pi
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O wow! I thought it was tan^2 x all along lol until Northstar wrote cot^2 x.